All Answers 356763 Answers

This problem is ill stated.

Unless you've left some info out there are infnite solutions for y in terms of x.

There end up being two free parameters, v0 and theta, the angle at which the throw is made with

respect to the x axis.

The ball could the thrown straight up and satisfy what's given or it could thrown at any angle and 

a v0 could be found to solve the conditions listed.

Did you leave something out?

Please use brackets if using square roots.

If I'm to interpert this as -30 + (5sqrt3)/35 

answer: (-210 + sqrt3)/7

If I'm to interpert this as (-30 + 5sqrt3)/35

answer: (-6 + sqrt3)/7

i've solved it all good!

I already answered you (A full day before you edited your question) 

You have not responded to my answer in any way.

What is wrong with you?  

Are you just totally rude?

I even read your mind and answered the question that you wanted to ask, rather than the garbled nonsense that you wrote.

yep the second choice was correct thanks so much

Second choice....just pick one to see if it fits

when j = 5      1/2^4 = 1/16    yep !

No problem it okay haha

Correct, EP   !!!

18 = 6 * 3  =  2* 3 * 3

93 = 31 * 3

31 and 2  are co-prime.....so.....GCF =  3

Need just a 'leetle' more info please..... 

The lines 2x+y=k and 0.5x+y=14 intersect when x=  -8.4. What is the value of k?

The intersection point will also have the same y value

So

-2x + k   =     -0.5x + 14        

-2(-8.4) + k = -.0.5(-8.4) + 14

16.8 + k =  4.2 + 14

16.8 + k =  18.2

k = 18.2 - 16.8   =   1.4

93 is what I think.

At x = - 8.4

.5(-8.4) + y = 14

y = 14 +4.2 = 18.2

2(-8.4) +18.2 = k = 1 .4

Looks like the difference between terms is  -48.8 -  (-35) = -48.8 + 35 =    -13.8

So.......

a_n = a_n-1 - 13.8 , where a₁ = - 7.4

 [ Remember that  a_n  is the term after a_n-1 ]

B

6          10

A           C

Cos A  =  AC / BC

By the Pythagorean Theorem

AC = sqrt ( BC^2 - AB^2)  = sqrt ( 10^2 - 6^2)  = sqrt (100 - 36)   = sqrt (64) = 8

So

Cos A =    8 / 10   =  4 / 5

2.  The diagram is not drawn to scale, but the measurements of the line segments and the right angles are correctly labeled. As shown in the diagram, BE=EC. Are the red and blue triangles similar? If so, enter the side of the blue triangle that corresponds to AB. If not, enter "no."

It's obvious that, if the triangles are similar, then ΔABE is similar to ΔECD

Note that....if similar, the scale factor is  120/35 = 24/7

Since...if similar, then DC =  (7/24)BE = (7/24)CE

Let CE = x   ....so DC = (7/24)x

Then, by the Pythagorean Theorem,

sqrt ( x^2 + (7x/24)^2 ] = 35

x *sqrt [ 576 + 49] / 24  =  35

x * sqrt (625) = 24 * 35

x * 25 =  840

x = 840 /25 =  33.6  =  CE  = BE

And...if similar..... AB = (24/7)CE

So....by the Pythagorean Theorem......

sqrt (BE^2 + AB^2 ) = 

sqrt (BE^2  + (24 CE/ 7)^2 ) =

sqrt ( 33.6^2 + [24 (33.6)/ 7]^2 )  =

sqrt ( 1128.96 + 13.271.04) =

sqrt ( 14400) =  120 = AE

So.....they are similar

AB  is similar to EC

3/7 x 1/2 =   3/14

(3/5 - 1/3) =  [ 9 - 5 ] / 15  =   4 / 15

So

(3/14)   /    (4/15)   =

(3/14) * ( 15/4)  =

45/ 46

numerator first:

3/7 *1/2 = 3/14

denominator:

3/5 - 1/3 = 9/15-5/15 (common denominator) = 4/15

3/14 divided by 4/15 is the same as multiplying by 15/4

3/14 *15/4 = 45/46 final answer

The first one is answered, here :

https://web2.0calc.com/questions/plz-help-and-explain-thoroughly-helpp

4. We know Angle A=45 degrees, Angle B=15 degrees and BC=sqrt(6). What is AB?

Angle C =  120°

Using the Law of Sines

BC / sin A   =  AB/ sin C

sqrt (6) /  (1/sqrt(2) ) =  AB / ( sqrt (3) / 2)

sqrt (12) = 2 AB / sqrt (3)

sqrt (3) * sqrt (12) = 2 AB

sqrt (36)  = 2 AB

6  =  2AB

AB =  3

thanks so much! 

I had to do that for a while, but then for some reason, it went away and I didn't have to push it anymore.........

.......except for the time I accidentally answered as a guest earlier today

No big deal.....easy to make mistakes....!!!!

We have the following number of possible sets of numbers.....we have six numbers and we want to choose any two of them....so....

C(6,2) = 15 sets

These are 

(1,2) (1,3) (1,4) (1, 5) (1,6)

(2, 3) (2,4) (2, 5) (2, 6)

(3,4) (3,5) (3,6)

(4, 5) ( 4, 6)

(5, 6)

Note, Ant, that only two of them have a product of six

So.....thirteen don't